Question

A study of the effect of television commercials on 12-year-old children measured their attention span, in seconds. The commercials were for clothes, food, and toys.

Clothes	Food	Toys
43	30	52
24	38	58
42	46	43
35	54	49
28	47	63
31	42	53
17	34	48
31	43	58
20	57	47
	47	51
	44	51
	54

Complete the ANOVA table. Use 0.05 significance level. (Round the SS and MS values to 1 decimal place and F value to 2 decimal places.)

Find the values of mean and standard deviation. (Round the mean and standard deviation values to 3 decimal places.)

Answer 1

Ans:

	Group 1	Group 2	Group 3	Total
Sum	271	536	573	1380
Count	9	12	11	32
Mean, Sum/n	30.111	44.667	52.091
Sum of square, Ʃ(xᵢ-x̅)²	648.889	722.667	326.909
Standard deviation	9.006	8.105	5.718

Number of treatment, k =	3
Total sample Size, N =	32
df(between) = k-1 =	2
df(within) = N-k =	29
df(total) = N-1 =	31
SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N =	2437.0
SS(within) = SS1 + SS2 + SS3 =	1698.5
SS(total) = SS(between) + SS(within) =	4135.5
MS(between) = SS(between)/df(between) =	1218.5
MS(within) = SS(within)/df(within) =	58.6
F = MS(between)/MS(within) =	20.81
p-value = F.DIST.RT(20.8053, 2, 29) =	0.0000

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	2437.0	2	1218.5	20.81	0.0000
Within Groups	1698.5	29	58.6
Total	4135.5	31

Null and Alternative Hypothesis:
Ho: µ1 = µ2 = µ3
H1: At least one mean is different.
Test statistic:
F =	20.81
Critical value:
Critical value Fc = F.INV.RT(0.05, 2, 29) =	3.328
p-value:
p-value = F.DIST.RT(20.8053, 2, 29) =	0.000
Decision:
P-value < α, Reject the null hypothesis.